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555-605 (Medium)|   Algebra|      
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Bunuel
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If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2 05 Apr 2016, 08:40
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E
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35% (medium)

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70% (01:49) correct 30% (02:32) wrong based on 238 sessions

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If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2 and then subtracting 2 times the reciprocal of xy produce which of the following?

A. \(x^2 + y^2 – 2xy\)

B. \(\frac{(x^2y^2 + xy)}{(x + y^2)}\)

C. \((\frac{1}{y}−\frac{1}{x})\)

D. \(\frac{(x^2 + y^2)}{(x^2y^2)}\)

E. \((\frac{1}{x} − \frac{1}{y})^2\)
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E
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If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2 05 Apr 2016, 09:32
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Bunuel
If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2 and then subtracting 2 times the reciprocal of xy produce which of the following?

A. x^2 + y^2 – 2xy
B. (x^2y^2 + xy)/(x + y^2)
C. (1/y−1/x)
D. (x^2 + y^2)/(x^2y^2)
E. (1/x − 1/y)^2
Show more


Hi,
The Q may seem complex but in actually is straight forward--
\(\frac{1}{x^2}+\frac{1}{y^2} -\frac{2}{xy}\)

TWO WAYS--


1) substitute
take x = 1/3 and y = 1/2..
\(3^2 +2^2-12\) = \(1\)
substitute values and see..

A. \(x^2 + y^2 – 2xy\)------- 1/9 +1/4 -1/3 -- NO
B. \(\frac{(x^2y^2 + xy)}{(x + y^2)}\)... (1/36 + 1/6)/(1/3+1/4) --NO
C. \((\frac{1}{y}−\frac{1}{x}).\).. 2-3= -1---NO
D. \(\frac{(x^2 + y^2)}{(x^2y^2)-}\)- (1/9+1/4)/(1/36)=13.. NO
E. \((\frac{1}{x}− \frac{1}{y})^2\).. (3-2)^2=1.. CORRECT

2) formula
VERY easy if you can see the hidden formula
\(\frac{1}{x^2}+\frac{1}{y^2} -\frac{2}{xy}\)..
\(\frac{1}{x}^2+\frac{1}{y}^2 - 2*\frac{1}{x}*\frac{1}{y}\)..
\((\frac{1}{x}− \frac{1}{y})^2\)..
same as E
ans E
General Discussion
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If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2 27 Jan 2017, 13:05
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Bunuel
If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2 and then subtracting 2 times the reciprocal of xy produce which of the following?

A. \(x^2 + y^2 – 2xy\)

B. \(\frac{(x^2y^2 + xy)}{(x + y^2)}\)

C. \((\frac{1}{y}−\frac{1}{x})\)

D. \(\frac{(x^2 + y^2)}{(x^2y^2)}\)

E. \((\frac{1}{x} − \frac{1}{y})^2\)
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Putting x=1, y=2 in \(\frac{1}{x^2} + \frac{1}{y^2} - \frac{2}{xy}\) we get \(\frac{1}{4}\)
Now TEST THE ANSWERS only E fits in.
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If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2 16 Sep 2020, 06:16
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Bunuel
If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2 and then subtracting 2 times the reciprocal of xy produce which of the following?

A. \(x^2 + y^2 – 2xy\)

B. \(\frac{(x^2y^2 + xy)}{(x + y^2)}\)

C. \((\frac{1}{y}−\frac{1}{x})\)

D. \(\frac{(x^2 + y^2)}{(x^2y^2)}\)

E. \((\frac{1}{x} − \frac{1}{y})^2\)
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Don't reciprocal first. Instead, just write x^2 + y^2 -2xy,
you got the formula (x-y)^2
now do the reciprocal.
The answer is E.
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If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2 16 Sep 2020, 07:00
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Bunuel
If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2 and then subtracting 2 times the reciprocal of xy produce which of the following?

A. \(x^2 + y^2 – 2xy\)

B. \(\frac{(x^2y^2 + xy)}{(x + y^2)}\)

C. \((\frac{1}{y}−\frac{1}{x})\)

D. \(\frac{(x^2 + y^2)}{(x^2y^2)}\)

E. \((\frac{1}{x} − \frac{1}{y})^2\)


Don't reciprocal first. Instead, just write x^2 + y^2 -2xy,
you got the formula (x-y)^2
now do the reciprocal.
The answer is E.
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May not be correct.
Reciprocal of \((x-y)^2=(\frac{1}{x-y})^2,\) which is not same as E.
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If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2 16 Sep 2020, 08:20
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Bunuel
If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2 and then subtracting 2 times the reciprocal of xy produce which of the following?

A. \(x^2 + y^2 – 2xy\)

B. \(\frac{(x^2y^2 + xy)}{(x + y^2)}\)

C. \((\frac{1}{y}−\frac{1}{x})\)

D. \(\frac{(x^2 + y^2)}{(x^2y^2)}\)

E. \((\frac{1}{x} − \frac{1}{y})^2\)
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The reciprocals: \(\frac{1}{x^2}\) and \(\frac{1}{y^2}\).

Adding the reciprocals: \(\frac{1}{x^2} + \frac{1}{y^2}\).

The reciprocal of xy: \(\frac{1}{xy}\).

Subtracting 2 times the reciprocal of xy: \(\frac{-2}{xy}\).

Putting everything together:

\(\frac{1}{x^2} + \frac{1}{y^2} - \frac{2}{xy}\)

\(= \frac{y^2 + x^2 - 2xy }{ x^2y^2}\)
(If you're sharp enough you can judge from here that the answer it is not ABCD so it has to be E.)

\(= \frac{(x - y)^2 }{ x^2y^2}\)

\(= (\frac{(x - y) }{ xy})^2\)

\(=(\frac{1}{y} - \frac{1}{x})^2 = (\frac{1}{x} - \frac{1}{y})^2\)

Ans: E
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If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2 16 Sep 2020, 10:29
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If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2 and then subtracting 2 times the reciprocal of xy produce which of the following?

A. \(x^2 + y^2 – 2xy\)

B. \(\frac{(x^2y^2 + xy)}{(x + y^2)}\)

C. \((\frac{1}{y}−\frac{1}{x})\)

D. \(\frac{(x^2 + y^2)}{(x^2y^2)}\)

E. \((\frac{1}{x} − \frac{1}{y})^2\) --> correct: \(\frac{1}{x^2}+\frac{1}{y^2}-\frac{1}{xy} = \frac{y^2+x^2-2xy}{x^2y^2} = \frac{(x-y)^2}{x^2y^2} = (\frac{1}{x}-\frac{1}{y})^2\)
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If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2 27 Oct 2025, 14:54
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